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thechocoluver445 Group Title

If the direction angle of vector a is 100 degrees, with a magnitude of 8, and the direction angle of vector c is 60 degrees, with a mangitude of 13, find the magnitude of resultant a + c. I know how to set this up and solve it, but for some reason I can't get the right answer.

  • one year ago
  • one year ago

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  1. thechocoluver445 Group Title
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    Here's what I did: |dw:1358034696496:dw| \[x = \sqrt{8^2 + 13^2 - 2(8)(13)(\cos 140°)}\]

    • one year ago
  2. zepdrix Group Title
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    Hmmm this is how I would do it :O So if we write our vectors in component form, we have,\[\large \vec{a}=8\cos100 \hat i+8\sin100 \hat j\]\[\large \vec c=13\cos60 \hat i+13\sin60 \hat j\]\[\large \vec a +\vec c=\color{#3366CF}{(8\cos100+13\cos60)\color{black}{\hat i}+(8\sin100+13\sin60)\color{black}{\hat j}}\] Hmm since 100 degrees isn't a nice special angle, we're going to get ugly decimal values it seems... So punch that into the calculator, then to get the MAGNITUDE of that vector we'll do uhhhh the thing, yah... I'm a little confused where your formula for x is coming from. It looks similar to the law of cosines. Maybe I just haven't done these types of problems in too long c: heh

    • one year ago
  3. thechocoluver445 Group Title
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    Yeah I used law of cosines. If I add up all of the things in your method, I don't get the correct answer.

    • one year ago
  4. zepdrix Group Title
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    ah ok c: lemme take another shot at it. sec.

    • one year ago
  5. zepdrix Group Title
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    Hmm I'm not quite sure :c In your attempt at the problem, where is the 140 degree angle coming from?

    • one year ago
  6. thechocoluver445 Group Title
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    180 - 40 = 140.

    • one year ago
  7. thechocoluver445 Group Title
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    (In the parallelogram)

    • one year ago
  8. mathmate Group Title
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    Did you have a positive or negative value for cosine?

    • one year ago
  9. mathmate Group Title
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    What did you get for the magnitude?

    • one year ago
  10. mathmate Group Title
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    @thechocoluver445

    • one year ago
  11. thechocoluver445 Group Title
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    It would be negative since the 140 is to the left of the y axis

    • one year ago
  12. mathmate Group Title
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    Sounds good. So your magnitude is greater than 13? What did you get? Is it a matter of number of decimals?

    • one year ago
  13. mathmate Group Title
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    @thechocoluver445 are you still there?

    • one year ago
  14. thechocoluver445 Group Title
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    It's supposed to be 9.8 but I keep getting 19. something!

    • one year ago
  15. thechocoluver445 Group Title
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    I'm pretty sure my way makes sense, but I don't know.

    • one year ago
  16. thechocoluver445 Group Title
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    I just realized that my way was correct. Since the resultant is opposite an obtuse angle, it must be the longest side of the triangle. Thus, 9.8 cannot be the right answer! So I was correct. Thanks to everyone for helping. :)

    • one year ago
  17. mathmate Group Title
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    19.8 is correct for the problem you have posted..

    • one year ago
  18. thechocoluver445 Group Title
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    The answer key was just wrong, lmao!

    • one year ago
  19. thechocoluver445 Group Title
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    Yeah, that's what I got lol!

    • one year ago
  20. thechocoluver445 Group Title
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    She probably just forgot the 1 before the 9.8 haha

    • one year ago
  21. mathmate Group Title
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    The only thing you may want to check to see if there are typos. It is possible that they modified the question but forgot to modify the key (which is half of the current answer).

    • one year ago
  22. mathmate Group Title
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    So go with 19.8.

    • one year ago
  23. mathmate Group Title
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    Oh, so it's not a printed key! That make it even more probably that the answer is wrong.

    • one year ago
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